1.

Mark the axymmetric carbon atoms and give the number of optical isomers in the following compounds: (i) CH_(3)-(CHOH)_(2)-COOH (ii) HOCH_(2)-(CHOH)_(4)-CHO (iii) HOCH_(2)-(CHOH)_(4)-CH_(2)OH

Answer»

Solution :
the compound cannot be divided into equal HALVES and it consists of two ASYMMETRIC carbon atoms
Hence, the NUMBER of d- and il-(optically active ) isomers,
`a=2^(n)=2^(2)=4`
and Number of meso forms, m=0
So, Total optical isomers `=a+m-4+0=4`


Where, `n=4,a=2((4-1))-2^(3)=8,m=2^((4//2-1))=2^((2-1))=2`.
`r=(8)/(2)=4`
So, Total optical isomers =8+2=10


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